Sum of first n natural numbers is (n) x (n + 1)/2 Here we have the sum = 100 x (101)/2 = 50 x 101 = 5050
The sum of the whole numbers from 1 to 100 inclusive is 5,050.
s100=50(2+99) this is because a=1, d=1 and n=100. answer is 5050
The sum of all the odd numbers from 1 through 100 is 10,000
n*(n+1)=sum 100*(100+1)=10100
The sum of the squares of the first 100 natural numbers [1..100] is 338350, while the sum of the first 100 natural numbers squared is 25502500.
100/3-1, 100/3 and 100/3+1 that is, 99, 100 and 101.
Sum of first n natural numbers is (n) x (n + 1)/2 Here we have the sum = 100 x (101)/2 = 50 x 101 = 5050
The sum of the numbers from 1 through 100 is 5,050.
That would be equal to the sum of all natural numbers less than or equal to 100 minus the sum of all natural numbers less than or equal to 89. The sum of all natural numbers less than or equal to x can be calculated with this equation: x ∑n = n(n + 1) / 2 n=1 We can take the numbers given above then, and plug them into that sum, giving us the equation: x = 100(100 + 1)/2 - 89(89 + 1) / 2 x = 5050 - 4005 x = 1045
The sum of the whole numbers from 1 to 100 inclusive is 5,050.
101
s100=50(2+99) this is because a=1, d=1 and n=100. answer is 5050
The sum of the all prime numbers from 1 to 100 is 1,161
The sum of all the odd numbers from 1 through 100 is 10,000
n*(n+1)=sum 100*(100+1)=10100
You add the numbers in a loop. Here is an example in Java:int sum = 0;for (int i = 1; i